Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Foundation
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- 1.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
- 2.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 3.A car's windscreen wiper blade is 40 cm long and sweeps through an angle of 110° as it wipes the glass. Using π = 3.14, work out the area of glass it wipes, to the nearest cm².
- 4.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
- 5.A kitchen scale is marked in equal divisions of 20 g. The pointer rests three divisions past the 400 g mark. Work out the reading on the scale.
- 6.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 7.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 8.A solid rubber ball has a radius of 3 cm. Work out the volume of the ball. Use π = 3.14.
- 9.A sector of a circle has an angle of 90° and an arc length of 15.7 cm. Using π = 3.14, work out the radius of the circle.
- 10.A rectangular garden bed measures 5 m by 3 m. A semicircular flower border is attached along one of the 3 m-wide ends, on the outside of the rectangle, so that the 3 m side is the diameter of the semicircle. Work out the total perimeter of the combined shape (the two long sides, the one remaining short side, and the curved edge). Use π = 3.14.
- 11.A vertical line passes through the point (5, 2). Write down the equation of this line.
- 12.A footpath is 3750 m long. Work out this distance in kilometres.
- 13.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
- 14.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
- 15.The top of a clock tower is 25 m above level ground. Oliver stands on the ground 25 m from the foot of the tower. Work out the angle of elevation of the top of the tower from the point where Oliver stands.
Answer key
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (d) 1535 cm² — The wiper sweeps out a sector of radius 40 cm, the blade length, through an angle of 110°. Sector area is angle ÷ 360 × π × radius²: 110 ÷ 360 × 3.14 × 1600 = 1535.1 cm², which rounds to 1535 cm². Forgetting to square the radius, using radius instead of radius², gives 38 cm². Using 110 ÷ 180 instead of 110 ÷ 360 for the fraction gives 3070 cm². Treating the 40 cm blade length as a diameter, so using a radius of 20 cm, gives 384 cm².
- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
- (c) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (a) 113.04 cm³ — Method: the volume of a sphere is (4 ÷ 3) × π × r³. Cube the radius, multiply by π, then multiply by 4 and divide by 3. Working: r³ = 3³ = 27, then 3.14 × 27 = 84.78, then 84.78 × 4 = 339.12 and 339.12 ÷ 3 = 113.04. Answer: 113.04 cm³. The distractors: 84.78 cm³ comes from stopping at πr³ and leaving out the four thirds; 37.68 cm³ comes from squaring the radius instead of cubing it, (4 ÷ 3) × 3.14 × 9; 28.26 cm³ comes from using πr², the area of a circle, and labelling it as a volume.
- (b) 10 cm — Arc length = (angle ÷ 360) × 2 × π × r, so 15.7 = (90 ÷ 360) × 2 × 3.14 × r = 1.57r, giving r = 15.7 ÷ 1.57 = 10 cm. (2.5 cm comes from forgetting the angle fraction and dividing by 2π alone; 20 cm comes from leaving out the factor of 2 in the arc length formula before dividing; 5 cm comes from dividing the arc length by π alone, ignoring both the angle fraction and the factor of 2.)
- (c) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (a) 3.75 km — To convert metres to kilometres, divide by 1000: 3750 ÷ 1000 = 3.75 km. Dividing by 100 instead of 1000 gives 37.5 km. Dividing by 10 instead of 1000 gives 375 km. Dividing by 10 000 instead of 1000 gives 0.375 km.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
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